Trying to find the right answer to this..

We have (x + a) (x + b) = x 2 + (a + b)x + ab Putting a = 2, b =3 and x = 4, we have LHS = (x + a) (x + b) = (4 + 2) (4 +3) = 6 x 7 = 42 RHS = x 2 + (a + b)x + ab = (4) 2 + ( 2 +3)x + (2) x (3) = 16 + 5 x 4 + 6 = 16 +26 = 42 i.e. LHS = RHS Thus, the given identity is true for the given values.

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Algebraic Expressions and Identities

Short Answer Type

151.

Subtract 4p²q + 5pq² – 3pq + 7q – 8p – 10 from 5p²q – 2pq² + 5pq – 11q – 3p + 28

188 Views

152.

Find the product of (5p² + 4p + 7) and (p + 2).

128 Views

153.

Simplify 3x(2x – 1) + 8 and find its values at x = –1.

115 Views

154.

Simplify: (x² – 6)(x – 6) + 36

95 Views

155.
Verify the identity (x + a)( x + b) = x² + (a + b)x + ab for a = 2, b = 3 and x = 4,

We have

(x + a) (x + b) = x²+ (a + b)x + ab

Putting a = 2, b =3 and x = 4, we have

LHS = (x + a) (x + b)
      = (4 + 2) (4 +3)
      = 6 x 7
      = 42

RHS = x²+ (a + b)x + ab
        = (4)² + ( 2 +3)x + (2) x (3)
        = 16 + 5 x 4 + 6
        = 16 +26
    = 42

     i.e.       LHS = RHS
Thus, the given identity is true for the given values.